What is maximum value of $f(t)=16\cos t \cdot \cos 2t \cdot\cos 3t \cdot \cos6t$ What is maximum value of $$f(t)=16\cos t \cdot \cos 2t \cdot\cos 3t \cdot \cos6t$$
My Approach:
$1. \;\;$Directly $t=0$ gives me maximum value of $16$.
$2.\;\;$ Converted it into $f(t)=\dfrac{\sin(4t)\sin(12t)}{\sin (t)\sin(3t)}\;\;$ but couldn't proceed further from this step.
My Doubts: $1.\;$Can we get maximum value without putting $t=0\;?$
$2.\;$ How can i proceed further using second method to get maximum value?
$3.\;$ Is there other way to solve this problem ?
 A: I think you did not answer the interesting part of the problem: why is $t=0$ a maximum of $f$? Wang YeFei did answer this.
To answer your doubts/questions:

*

*You cannot get the maximum value without evaluating the function at a maximum.

*Your conversion of $f$ is not even defined at $t=0$. So this way can not work.

*The usual way would be to find $f'$ to get all local extrema and pick the one with highest value as global maximum. But this would be unnecessaryly complicated here…

A: Observe that $\forall t, f(t) \le 16$ and $f(t) = 16$ when $t = 0$. This shows $16$ is the max value of $f$. All you need is $\cos(kt) \le 1$ for any $k,t$.
A: Answering your question...

*

*Yes, but it would involve setting $f'(t)=0$ - you can use logarithmic differentiation$^1$ to get the derivative or use the product rule multiple times.


*Same as 1, but now you'd have to involve the quotient rule AND the product rule to get things moving.


*You could graph the equation and see where the maxima are by inspection. Desmos or Geogebra would be a huge help here.
$^1$ To find the logarithmic derivative, we would have...
$$f(t) = 16 \cos t \cos 2t \cos 3t \cos 6t \\ \ln f(t) = \ln 16 + \ln \cos t + \ln \cos 2t + \ln \cos 3t + \ln \cos 6t \\ \frac {f'(t)}{f(t)} = 0 + \frac {1}{\cos t}(-\sin t) + \frac {1}{\cos 2t}(-2\sin 2t) + \frac {1}{\cos 3t}(-3\sin 3t) + \frac {1}{\cos 6t}(-6\sin t) \\ f'(t) = f(t) (-\tan t -2 \tan 2t - 3 \tan 3t - 6 \tan 6t) \\ f'(t) = 16 \cos t \cos 2t \cos 3t \cos 6t(-(\tan t +2 \tan 2t + 3 \tan 3t + 6 \tan 6t))$$  Setting $f'(t) = 0$, you would have either $$16 \cos t \cos 2t \cos 3t \cos 6t = 0$$ or $$-(\tan t +2 \tan 2t + 3 \tan 3t + 6 \tan 6t)= 0$$
